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1 câu trả lời 2
Malic acid có công thức cấu tạo: \({\rm{HOOC}} - {\rm{CH}}({\rm{OH}}) - {\rm{C}}{{\rm{H}}_2} - {\rm{COOH}}\)
\(\begin{array}{l}{\rm{HOOC}} - {\rm{CH}}({\rm{OH}}) - {\rm{C}}{{\rm{H}}_2} - {\rm{COOH}} + 3{\rm{Na}} \to {\rm{NaOOC}} - {\rm{CH}}({\rm{ONa}}) - {\rm{C}}{{\rm{H}}_2} - {\rm{COONa}} + 1,5{{\rm{H}}_2}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1,5a\,\,mol\end{array}\)
\(\begin{array}{l}{\rm{HOOC}} - \mathop {\rm{C}}\limits_{|\,} {\rm{H}} - {\rm{C}}{{\rm{H}}_2} - {\rm{COOH}} + 2{\rm{NaHC}}{{\rm{O}}_3} \to {\rm{NaOOC}} - \mathop {\rm{C}}\limits_| {\rm{H}} - {\rm{C}}{{\rm{H}}_2} - {\rm{COONa}} + 2{\rm{C}}{{\rm{O}}_2} + 2{{\rm{H}}_2}{\rm{O}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,OH\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,OH\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{a}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{2a}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{mol}}\end{array}\)
\( \Rightarrow \left\{ {\begin{array}{*{20}{l}}{{{\rm{n}}_{{{\rm{H}}_2}}} = 1,5{\rm{a}}}\\{{{\rm{n}}_{{\rm{C}}{{\rm{O}}_2}}} = 2{\rm{a}}}\end{array} \Rightarrow \frac{{{{\rm{V}}_1}}}{{\;{{\rm{V}}_2}}} = \frac{{1,5}}{2} \Rightarrow 4\;{{\rm{V}}_1} = 3\;{{\rm{V}}_2}} \right.\).
Chọn B.
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